A key question is: what causes the electrons to flow from the anode to the cathode?

The electrons flow from the anode to the cathode because of a difference in the potential energy of the electrons at the anode compared to the cathode

In particular, the potential energy of the electrons is higher at the anode than at the cathode

Like water molecules in a waterfall, the electrons move in a direction from high potential energy to low potential energy (i.e. "downhill" in an energy landscape)

The potential difference between two electrodes is measured in Volts (V)

One volt is the potential difference, between an anode and a cathode, that would be required in order to impart 1 Joule of energy to a charge of 1 coulomb (note: 1 coulomb is the charge of a collection of 6.24 x 1018 electrons)

1 V = 1 J/C (i.e. energy per electron)

The potential difference between two electrodes is called the electromotive force, or emf

The emf of a voltaic cell is called the cell potential, or the cell voltage

The cell voltage of a voltaic cell will be positive value

The cell voltage will depend upon the specific half-reactions, concentrations of ions, and temperature

Common tabulations of cell voltages are performed at 25°C and standard conditions of concentrations of liquid (i.e. 1M) and gas components (i.e. 1 atm)

Under standard conditions, the emf of a voltaic cell is called the standard emf or the standard cell potential (E0cell)

By convention, the potential associate with an electrode is the potential for reduction

The standard cell potential is given by the standard reduction potential of the cathode, minus the standard reduction potential of the anode

E0cell = E0red(cathode) - E0red(anode)

If E0red(cathode) > E0red(anode) then the electron flow from anode to cathode is spontaneous

Standard Reduction Potentials

The cell potential of a voltaic cell depends upon the two redox half-reactions that comprise the functioning cell

There are many different combinations of redox half-reactions. Do we need to tabulate all possible unique combinations?

As you might have guessed, the answer is no, we don't. In principle, we could assign a unique standard potential to each half-reaction and by comparing these values determine the cell potential for any combination of half-reactions

There is one problem, however: we have seen that an isolated half-reaction doesn't result in reduction or oxidation or electron flow. Therefore, how are we going to assign a standard potential to a half-reaction?

We are going to use a generally agreed upon reduction reaction as the reference half-reaction against which to compare and tabulate all possible half-reactions

The reference reaction is the reduction of H+(aq) ions to produce H2(g):

2H+(aq, 1M) + 2e-® H2(g, 1 atm) E0red = 0.0V

This is a Standard Hydrogen Electrode (SHE)

How do we use this SHE to determine the standard reduction potential, E0red, for some redox half-reaction?

Let's say we are interested in determining the standard reduction potential for Zn2+

We would set up a voltaic cell using a SHE as one half-cell, and a Zinc electrode as the other half-cell

The half-cell reactions are:

Zn(s) ® Zn2+(aq) + 2e- (Zinc electrode)

2H+(aq) + 2e-® H2(g) (SHE)

The overall cell reaction:

Zn(s) + 2H+(aq) ® Zn2+(aq) + H2(g)

Note: this is performed under Standard Conditions, where concentrations = 1.0M, and p = 1 atm

For the reaction as shown, the zinc electrode is the anode (oxidation), and the SHE is the cathode (reduction)

The voltage for the overall cell is measured at 0.76 Volts (i.e. this is the standard cell potential)

From the previous definition of the standard cell potential:

E0cell = E0red(cathode) - E0red(anode)

We know the value for the standard cell potential, it is +0.76V. We also know the value for the reduction potential for the SHE, it is 0V. Thus,

0.76V = E0red(cathode) - E0red(anode)

0.76V = 0 - E0red(anode)

E0red(anode) = -0.76V

In other words, in reference to the SHE, the standard reduction potential for the zinc electrode is determined to be -0.76V.

This may seem a little strange because the cell reads +0.76V and yet the determined reduction potential for the zinc electrode is -0.76V.

The key here is that the original definition of the standard cell potential is set up such that each cell is referenced as a reduction reaction.

In other words what we have actually determined is the half-cell potential (in reference to the SHE) for the reduction of zinc:

Zn2+(aq) + 2e-® Zn(s) E0red = -0.76V

Whenever we assign a potential to a half-reaction using the equation for the standard cell potential, it is always done in reference to a reduction reaction

Thus, in reference to the SHE, the oxidation of Zn(s) is associated with a half-cell potential of +0.76V (this is the spontaneous direction for the reaction; the reduction of Zn2+ and the oxidation of H2(g) is not spontaneous)

The standard cell potential (Voltage) describes the potential per unit charge that the reaction produces

This value of the potential per unit charge (i.e. electron) is a constant that is a property of the two half-reactions that are reacting in the cell. It is independent of the stoichiometry or amount of the reactants in the cell

Some standard reduction potentials at 25°C:

Standard Potential, (E0red) in Volts

Reduction Half-Reaction

0.80

Ag+(aq) + e-® Ag(s)

0.34

Cu2+(aq) + 2e-® Cu(s)

0

2H+(aq) + 2e-® H2(g)

-0.76

Zn2+(aq) + 2e-® Zn(s)

-3.05

Li+(aq) + e-® Li(s)

The more positive the value of E0redthe greater the driving force for reduction

In reference to H+/H2(g) , Ag+ will preferentially be reduced to Ag(s) and H2(g) will be oxidized to H+(aq). In reference to H+/H2(g) , Zn(s) will preferentially be oxidized and H+ will be reduced to H2(g).

The cathode is the electrode where reduction occurs, therefore, in order for a cell to operate spontaneously, the half-reaction at the cathode must have a more positive value of E0redthan the half-reaction at the anode

The greater driving force for reduction at the cathode can force the reaction at the anode to operate "in reverse" - i.e. to undergo the oxidation reaction

The difference between the standard reduction potentials of the two reactions, E0red (cathode) - E0red (anode), is the excess potential that can be used to drive electrons through the cell, E0cell

A silver-lithium battery is a voltaic cell with a silver cathode and a lithium anode. What is the voltage of this voltaic cell?

E0cell = E0red(cathode) - E0red(anode)

E0cell = 0.8V - (-3.05V) = 3.85 Volts

What will happen if a science project on batteries suggests that a cathode be made out of Zinc and the anode of Copper?

E0cell = E0red(cathode) - E0red(anode)

E0cell = -0.76V - (0.34V) = -1.10 Volts

The battery will run "backwards". In other words, the greater reduction potential of Copper will actually drive oxidation of the Zinc. If a volt meter is connected to the cell with the assumption that the Zinc is the cathode, then the voltage will read negative 1.10 volts

Oxidizing and Reducing Agents

In the above discussion of reduction potentials and their relationship to one another in half-reactions, we can predict which direction current will flow in a voltaic cell. In other words, we can predict which reactant will be oxidized and which will be reduced

The same information can be used for redox reactions where the compounds are in direct contact, and not separated into half-cells.

For example, from the example above, we would predict that silver ion will oxidize lithium metal as a spontaneous redox reaction. Alternatively, we would predict that the reduction of lithium ion by silver metal is not a spontaneous reaction (and would require the input of energy to drive)

The more positive the E0red value of a half-reaction, the greater the tendency of the compound to be reduced - and therefore to oxidize another compound (i.e. the greater the tendency to act as an oxidizing agent)